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Number of terms of A245630 <= n.
3

%I #22 May 08 2021 23:39:14

%S 1,1,1,1,1,2,2,2,2,2,2,2,2,2,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,3,

%T 4,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,5,

%U 5,5,5,5,5,5,5,5,6,6,6,6,6,6,6,6,6,6,6,6,6,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7,7

%N Number of terms of A245630 <= n.

%H Robert Israel, <a href="/A245636/b245636.txt">Table of n, a(n) for n = 1..10000</a>

%H Paul Erdős, <a href="http://www.jstor.org/stable/2306529">Solution to Advanced Problem 4413</a>, American Mathematical Monthly, 59 (1952) 259-261.

%F As n -> infinity, a(n)/sqrt(n) -> Product_{i=1..infinity} (1 - 1/prime(i))/(1 - (prime(i)*prime(i+1))^(-1/2)), see Erdős reference.

%e The first two terms of A245630 are 1 and 6, so a(n) = 1 for 1 <= n <= 5 and a(6) = 2.

%p N:= 10^4: # to get a(1) to a(N)

%p PP:= [seq(ithprime(i)*ithprime(i+1), i=1.. numtheory[pi](floor(sqrt(N)))-1)]:

%p ext:= (x, p) -> seq(x*p^i, i=0..floor(log[p](N/x))):

%p S:= {1}: for i from 1 to nops(PP) do S:= map(ext, S, PP[i]) od:

%p E:= Array(1..N):

%p for s in S do E[s]:= 1 od:

%p A:= map(round,Statistics:-CumulativeSum(E)):

%p seq(A(i),i=1..N);

%t M = 10^4;

%t T = Table[Prime[n] Prime[n+1], {n, 1, PrimePi[Sqrt[M]]}];

%t T2 = Select[Join[T, T^2], # <= M&];

%t S = Join[{1}, T2 //. {a___, b_, c___, d_, e___} /; b*d <= M && FreeQ[{a, b, c, d, e}, b*d] :> Sort[{a, b, c, d, e, b*d}]];

%t ee = Table[0, {M}];

%t Scan[Set[ee[[#]], 1]&, S];

%t Accumulate[ee] (* _Jean-François Alcover_, Apr 17 2019 *)

%Y Cf. A006094, A028260, A245630.

%K nonn

%O 1,6

%A _Robert Israel_, Jul 28 2014